Which one has the highest boiling point

Options

A. `0.1 N Na_(2)SO_(4)`
B. `0.1 N MgSO_(4)`
C. `0.1 M Al_(2)(SO_(4))_(3)`
D. `0.1 M BaSO_(4)`

Explanation

To determine which solution has the highest boiling point among the given options, we will use the concept of colligative properties, specifically the elevation in boiling point. The elevation in boiling point (\( \Delta T_b \)) is given by the formula:

\[ \Delta T_b = i \cdot K_b \cdot m \]

Where:
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( K_b \) = molal boiling point elevation constant (a property of the solvent)
- \( m \) = molality of the solution

Since we are dealing with dilute solutions, we can assume that molality (\( m \)) is approximately equal to molarity.

### Step-by-step Solution:

1. **Identify the solutions and their normalities/molarities**:
- Option 1: 0.1 N \( \text{Na}_2\text{SO}_4 \)
- Option 2: 0.1 N \( \text{MgSO}_4 \)
- Option 3: 0.1 M \( \text{Al}_2\text{(SO}_4\text{)}_3 \)
- Option 4: 0.1 M \( \text{BaSO}_4 \)

2. **Convert normality to molarity for the first two options**:
- For \( \text{Na}_2\text{SO}_4 \):
- Normality = 0.1 N, n-factor = 2 (since it dissociates into 2 Na\(^+\) and 1 SO\(_4^{2-}\))
- Molarity = Normality / n-factor = \( 0.1 / 2 = 0.05 \) M
- For \( \text{MgSO}_4 \):
- Normality = 0.1 N, n-factor = 2
- Molarity = \( 0.1 / 2 = 0.05 \) M

3. **Calculate \( i \) and \( i \cdot m \) for each solution**:
- **Option 1: \( \text{Na}_2\text{SO}_4 \)**:
- \( i = 3 \) (2 Na\(^+\) + 1 SO\(_4^{2-}\))
- \( m = 0.05 \)
- \( i \cdot m = 3 \cdot 0.05 = 0.15 \)

- **Option 2: \( \text{MgSO}_4 \)**:
- \( i = 2 \) (1 Mg\(^{2+}\) + 1 SO\(_4^{2-}\))
- \( m = 0.05 \)
- \( i \cdot m = 2 \cdot 0.05 = 0.1 \)

- **Option 3: \( \text{Al}_2\text{(SO}_4\text{)}_3 \)**:
- \( i = 5 \) (2 Al\(^{3+}\) + 3 SO\(_4^{2-}\))
- \( m = 0.1 \)
- \( i \cdot m = 5 \cdot 0.1 = 0.5 \)

- **Option 4: \( \text{BaSO}_4 \)**:
- \( i = 2 \) (1 Ba\(^{2+}\) + 1 SO\(_4^{2-}\))
- \( m = 0.1 \)
- \( i \cdot m = 2 \cdot 0.1 = 0.2 \)

4. **Compare the values of \( i \cdot m \)**:
- Option 1: \( 0.15 \)
- Option 2: \( 0.1 \)
- Option 3: \( 0.5 \)
- Option 4: \( 0.2 \)

5. **Determine the highest boiling point**:
- The solution with the highest \( i \cdot m \) value is Option 3: \( \text{Al}_2\text{(SO}_4\text{)}_3 \) with a value of \( 0.5 \).

### Conclusion:
The solution with the highest boiling point is **Option 3: 0.1 M \( \text{Al}_2\text{(SO}_4\text{)}_3 \)**.